Question

12. Two bulbs of 100 W each and two coolers of 250 W each, work on an average 6 hours a day. If the energy
costs Rs. 1-75 per kWh, calculate the monthly bill and the minimum fuse rating when power is supplied
at 250 V.
bhaiyo aur unki beheno iska and bata do pls​

Answer 1

Answer:

the monthly bill and the maximum fuse rating when power is supplied at 250v is Rs 6.475

Explanation:

solution:

we know that,

power p =energy/time

energy=power/time

2 bulbs of 100 w each,

100 ₓ 2 = 200 w per day

two coolers of 250 w each,

250 ₓ 2 = 500 w per day ,

The bulbs are working nearly 6 hours per day

so, the energy consumed in 1 day

for bulb = 200 ₓ 6 = 1200 key word optimized for developer = 1.2 KW/h

For cooler =500 ₓ 5 = 2500 KW/h =2.5 KW/h

Total energy consumed (add bulb and coolers watt)=3.7 KW/h

1 unit = 1 KW/h

1 unit costs Rs 1.75

on calculating 3.7 units,we get

3.7 ₓ 1.75 =Rs 6.475 (monthly bill